tożsamosc trygonometryczna

[Bartosz89M]

Sprawdz czy podana równość jest tożsamością trygonometryczną

a)\(\displaystyle{ (sinx+cosx) ^{2} + (sinx-cosx) ^{2}=2}\)

b)\(\displaystyle{ cos ^{4}x - sin ^{4}x=cos ^{2}x-sin ^{2}x}\)

c)\(\displaystyle{ (1+ tg ^{2}x) cos ^{2}x=1}\)

d)\(\displaystyle{ \frac{tgx+tgy}{ctgx+ctgy}=tgx tgy}\)

e)\(\displaystyle{ \frac{cos ^{x}-1 }{sin ^{2}x-1 }=tg ^{2}x}\)

f) \(\displaystyle{ \frac{sinx}{1+cosx}+ \frac{1+cosx}{sinx}= \frac{2}{sinx}}\)

g) \(\displaystyle{ \frac{tgx}{tgx+ctgx}=sin ^{2}x}\)

[natkoza]

chyba trygonometryczną
a)
\(\displaystyle{ L=(sinx+cosx)^2+(cosx-sinx)^2=sin^2x+2sinxcosx+cos^2x+sin^2x-2sinxcosx+cos^2x=2(sin^2x+cos^2)=2=P}\)

[Piotrek89]

b)

\(\displaystyle{ L=\cos ^{4}x-\sin ^{4}x=(\cos^{2}x-\sin ^{2}x)(\cos ^{2}x +\sin ^{2}x)=P}\)

[olcia_]

a). \(\displaystyle{ (sinx+cox) ^{2} +(sinx-cosx) ^{2} =2}\)
\(\displaystyle{ sin ^{2}x +2sinxcosx+cos ^{2}x +sin ^{2}x -2sinxcosx+cos ^{2}x =2}\)
\(\displaystyle{ 2(sin^{2}x +cos ^{2}x )=2}\)
\(\displaystyle{ 2=2}\)
b).\(\displaystyle{ (cos ^{2}x -sin ^{2} x)(cos ^{2} x+sin ^{2} x)=cos ^{2} x-sin ^{2} x}\)
\(\displaystyle{ cos ^{2} x+sin ^{2} x=1}\)
\(\displaystyle{ 1=1}\)
c). \(\displaystyle{ (1+( \frac{sinx}{cosx} )^{2} )*cos ^{2} x=1}\)
\(\displaystyle{ cos ^{2}x+sin ^{2} x=1}\)
\(\displaystyle{ 1=1}\)

[ Dodano: 28 Marca 2008, 19:33 ]
g). \(\displaystyle{ \frac{\frac{sinx}{cosx} }{\frac{sinx}{cosx} + \frac{cosx}{sinx} } =sin ^{2} x}\)
\(\displaystyle{ \frac{\frac{sinx}{cosx}}{ \frac{sin ^{2}x+cos ^{2} x }{sinxcosx} }=sin ^{2} x}\)
\(\displaystyle{ \frac{\frac{sinx}{cosx}}{ \frac{1}{sinxcosx} } =sin ^{2} x}\)
\(\displaystyle{ sinx*sinxcosx=sin ^{2} x}\)
\(\displaystyle{ sin ^{2} x=sin ^{2} x}\)

[ Dodano: 28 Marca 2008, 19:40 ]
f). \(\displaystyle{ \frac{sin ^{2} x+(1+cosx) ^{2} }{(1+cosx)sinx} = \frac{2}{sinx}}\)
\(\displaystyle{ \frac{sin ^{2}x+1+2cosx+cos ^{2} x }{(1+cosx)sinx}= \frac{2}{sinx}}\)
\(\displaystyle{ \frac{2+2cosx}{(1+cosx)sinx} = \frac{2}{sinx}}\)
\(\displaystyle{ \frac{2(1+cosx)}{(1+cosx)sinx} = \frac{2}{sinx}}\)
\(\displaystyle{ \frac{2}{sinx} = \frac{2}{sinx}}\)

[fala19]

g) L=\(\displaystyle{ frac{\(\displaystyle{ \frac{sinx}{cosx}}\) }{\(\displaystyle{ \frac{sinx}{cosx}}\) +\(\displaystyle{ \frac{cosx}{sinx}}\) }}\) = \(\displaystyle{ frac{\(\displaystyle{ \frac{sinx}{cosx}}\) }{\(\displaystyle{ \frac{1}{sinxcosx}}\) }}\) = \(\displaystyle{ \frac{sinx}{cosx}}\) * \(\displaystyle{ \frac{sinxcosx}{1}}\) =sin2x=P

[Bartosz89M]

dziekuje. a jak ten przyklad?

\(\displaystyle{ \frac{sinx}{1+cosx}+ \frac{1+cosx}{sinx}= \frac{2}{sinx}}\)

[tkrass]

\(\displaystyle{ \frac{sinx}{1+cosx}+ \frac{1+cosx}{sinx}=\frac{sin ^{2} x + (1+cosx) ^{2} }{(1+cosx)sinx}=\frac{sin ^{2} x+ cos ^{2} x + 1+2cosx }{(1+cosx)sinx} = \frac{2+2cosx}{(1+cosx)sinx}= \frac{2(1+cosx)}{sinx(1+cosx)}= \frac{2}{sinx}}\)

[Bartosz89M]

dzieki. juz wszystkie procz tego, nie mam pojecia:

\(\displaystyle{ \frac{tgx+tgy}{ctgx+ctgy}=tgx tgy}\)